题意:给定\(a[1...n]\),\(Q\)次询问求\(A[L...R]\)的异或组合再或上\(K\)的最大值
目前提交处于TLE状态,原因待查
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int MAXN = 1e4+11;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
ll a[MAXN];
struct LB{
ll b[33];
void clear(){
rep(i,0,31) b[i]=0;
}
void insert(int x){
rrep(i,31,0) if(x>>i&1){
if(b[i]) x^=b[i];
else{
b[i]=x;
//rrep(j,i-1,0) if(b[j]&&(b[i]>>j&1)) b[i]^=b[j];
//rep(j,i+1,30) if(b[j]>>i&1) b[j]^=b[i];
break;
}
}
}
ll rnk1(){
ll res=0;
rrep(i,31,0) res=max(res,res^b[i]);
return res;
}
};
ll best;
struct ST{
LB b[MAXN<<2],ans;
#define lc o<<1
#define rc o<<1|1
void merge(LB &x,LB &y){
rep(i,0,31) if(y.b[i]) x.insert(y.b[i]);
}
void pu(int o){
merge(b[o],b[lc]);
merge(b[o],b[rc]);
}
void init(){memset(b,0,sizeof b);}
void build(int o,int l,int r){
if(l==r){
if(a[l]&best) b[o].insert(a[l]&best);
return;
}
int mid=l+r>>1;
build(lc,l,mid);
build(rc,mid+1,r);
pu(o);
}
void query(int o,int l,int r,int L,int R){
if(L<=l&&r<=R){
merge(ans,b[o]);
return;
}
int mid=l+r>>1;
if(L<=mid) query(lc,l,mid,L,R);
if(R>mid) query(rc,mid+1,r,L,R);
}
}st;
int main(){
int T=read();
while(T--){
int n=read();
int q=read();
ll k=read();
best=0;
rep(i,0,31) if(!(k>>i&1)) best|=(1<<i);
rep(i,1,n) a[i]=read();
st.init();
st.build(1,1,n);
rep(i,1,q){
int L=read();
int R=read();
st.ans.clear();
st.query(1,1,n,L,R);
println(st.ans.rnk1()|k);
}
}
return 0;
}
原文地址:https://www.cnblogs.com/caturra/p/8976513.html
时间: 2024-10-03 16:52:59